Question

The volume occupied by 9.03×10^23 molecules of N2 gas at STP is closest to??
A) 1.50 L
B) .500 L
C) 22.4 L
D) 33.6 L

Answer 1

Answer:

D) 33.6 l

Explanation:

First convert from molecules of N2 gas to moles of N2 gas:

9.03 x 10^23 molecules N2 x (1 mole / 6.022x10^23 molecules) = 1.499501827 mol N2

Next,

Use the ideal gas equation:

PV = nRT

At STP conditions,

P = 1 atm

V = x

n = 1.499501827 mol N2

R = 0.08206 atm L / mol K

T = 273.15 K

Solve for V

PV = nRT

V = nRT / P = 33.6108671 L = 33.6 L

Answer 2

The volume occupied by 9.03×10²³ molecules of N₂ gas at STP is closest to 33.6 L

The correct answer to the question is Option D. 33.6 L.

We'll begin by calculating the number of mole of N₂ gas that contains 9.03×10²³ molecules. This can be obtained as follow:

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of N₂

Therefore,

9.03×10²³ molecules = $\frac{9.03*10^{23}}{6.02*10^{23}}$

9.03×10²³ molecules = 1.5 mole of N₂

Thus, 9.03×10²³ molecules is present in 1.5 mole of N₂.

Finally, we shall determine the volume occupied by 1.5 moles of N₂ at stp. This is illustrated below:

At standard temperature and pressure (STP),

1 mole of N₂ = 22.4 L

Therefore,

1.5 moles of N₂ = 1.5 × 22.4

1.5 moles of N₂ = 33.6 L

Thus, we can conclude that the volume occupied by 9.03×10²³ molecules of N₂ gas at STP is closest to 33.6 L

Option D. 33.6 L gives the correct answer to the question.

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